91Ó°ÊÓ

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton but has an approximate normal distribution. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the milk in all such cartons is always equal to 15 ounce. The quality control inspector at this company takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and tests the null hypothesis, \(\mu=32\) ounces, against the altemative hypothesis, \(\mu \neq 32\) ounces. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.93\) ounces. a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and adjust it if she chooses the maximum probability of a Type I error to be .01? What if the maximum probability of a Type I error is 05 ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\). Does the machine need to be adjusted? What if \(a=.05 ?\)

Step by step solution

01

Calculate the Test Statistic

First, calculate the Z value (also known as the test statistic) using the formula \[Z = {(X - \mu) }/{(\sigma / \sqrt{N})} \] where \(X\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(N\) is the size of the sample. Substituting in the given values, the test statistic \(Z\) comes out to be \[-0.2333\]

02

Calculate the P-value

The P-value is calculated as \[P(Z < -0.2333) \times 2\] because it's a two-tailed test. Looking up in the Z table, we find the P value which is found to be approximately \(0.8155\) for \(Z = -0.2333\). Therefore, the P-value is \(2 \times 0.8155 = 1.631\)

03

Decision Making with P-value and Type I Error

Firstly, it should be noted that the p-value cannot be greater than 1. Hence, the calculated p-value of 1.631 is incorrect. It seems there has been a miscalculation in step 2 while finding the p-value. The corrected p-value should be \(2 \times (1 - 0.8155) = 0.369\). Now, if the maximum probability of a Type I error is 0.01 or 0.05, both are less than the calculated p-value of \(0.369\). Therefore, the null hypothesis is not rejected. This means the quality control inspector will not decide to stop the machine and adjust it.

04

Critical-value Approach

For \(\alpha = .01\), the rejection region for a two-tailed test lies in \(Z > 2.57\) or \(Z < -2.57\). The calculated Z score of \(-0.2333\) does not lie in this region, so the null hypothesis is not rejected. The same conclusion can be drawn for \(\alpha = .05\), where the rejection region lies in \(Z > 1.96\) or \(Z < -1.96\). Therefore, in both scenarios, the machine doesn't need to be adjusted.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution

Understanding normal distribution is key to analyzing data in many scientific fields. It is a continuous probability distribution characterized by a symmetrical bell-shaped curve. This curve helps in making decisions as it provides insight into how the data is dispersed or varies.
Normally distributed data has some distinct features:

• The mean, median, and mode are all equal.
• It is symmetrical about the mean.
• Approximately 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.

In the milk carton exercise, the distribution of milk amounts follows this pattern. By understanding the properties of normal distribution, decisions on whether the machine is functioning correctly can be made more reliably.

P-value

The p-value is a powerful tool in hypothesis testing. It quantifies the probability of obtaining the observed sample results, or something more extreme, if the null hypothesis is true. In simpler terms, the smaller the p-value, the stronger the evidence against the null hypothesis.
In hypothesis testing, the p-value helps decide whether to accept or reject the null hypothesis:

• If the p-value is low (typically less than 0.05), the null hypothesis is rejected.
• If it's high, we fail to reject the null hypothesis.

For Farmer's Dairy exercise, the calculated p-value suggested that the machine was filling the cartons correctly, as it was higher than the maximum allowable probability for a Type I error.

Type I Error

A Type I error occurs when the null hypothesis is incorrectly rejected when it is true. It is also known as a "false positive." The probability of making a Type I error is denoted by the alpha level, \( \alpha \). Setting this level is crucial as it reflects how much risk one is willing to take in incorrectly rejecting the null hypothesis.
In the Farmer's Dairy context:

• An incorrect rejection could lead to unnecessary adjustments to the machine.
• Choosing an alpha level of 0.01 or 0.05 indicates the tolerance for making this kind of error.

By understanding and setting the desired alpha level, the quality control process can more effectively prevent unnecessary stoppages and adjustments.

Standard Deviation

Standard deviation is a measure of how spread out the values in a dataset are. It gives us insight into the variability or dispersion of the dataset. A low standard deviation close to zero indicates that the data points are very close to the mean, while a high standard deviation indicates more spread out data.
In the context of Farmer's Dairy, understanding the standard deviation of milk amounts can show how much variation exists in the filling process:

• If the standard deviation is small, it indicates a very consistent filling process.
• If large, it suggests variability that might need attention.

Calculating and interpreting standard deviation is crucial in process control to maintain product quality and consistency.